Find vectors when added up equal (1, 1, 1) Question:
Let $V$ be the 2-dim subspace of $\mathbb R^3$ spanned by $(1, 2, -3)$ and $(-2, 0, 1)$. Write the vector $u = (1,1,1)$ in the form $u = v + w$, where $v$ is in $V$ and $w$ is in $V^\perp$, which is the subspace orthogonal to $V$.
What I tried:
So I found a vector that was linearly independent to the two vectors given: $(0, 0, 1)$. Then I used the Gram–Schmidt process to find a vector orthogonal to both of them. I found $(2, 5, 4)$ is orthogonal to $(1, 2, -3)$ and $(-2, 0, 1)$. However, after obtaining the unit vectors, I don't know how to proceed. No matter what vectors I add up, I do not get $u$.
My unit vectors were:
$$u_1 = (1,2,-3)/\sqrt14$$
$$u_2 = (23,-10,1)/(3\sqrt(70)$$
$$u_3 = (2,5,4)/(3\sqrt5)$$
Please tell me how to obtain the vectors so when I add $v+w$, it will equal $(1,1,1)$. Thank you!
 A: These two equations should be enough:
1) The sum $w+v=(1,1,1)$
2) $u.v=0$ , where $.$ is the dot product.
Equivalently, $v=\frac {<(1,1,1),(1,2,-3)>}{||(1,2,-3)||}(1,2,-3)+\frac{<(1,1,1),(-2,0,1)>}{||(-2,0,1)||}(-2,0,1))$ , where $<,>$ is the standard dot product in $\mathbb R^3$ and $||v||$ is the standard (square of the) length given by $<v,v>$. 
This will give you the projection onto the span of {$(1,2,-3),(-2,0,1)$}. Then you get the vector along the other direction by subtracting the result above from $(1,1,1)$.
The general formula for projecting $v$ onto the subspace spanned by {$u_1, u_2$} is $$ \frac {<v,u_1>}{<u_1, u_1>}u_1 + \frac {<v,u_2>}{<u_2, u_2>}u_2 $$
A: You have found a basis. Let these vectors be the columns of a matrix $A$, and solve $Ax=(1,1,1)^T $.

Let $A = \begin{pmatrix} 2 & 1 & -2 \\ 5 & 2 & 0 \\ 4 & -3 & 1 \end{pmatrix}$. 
Compute $$Ax = \begin{pmatrix} 1 \\ 1\\ 1\end{pmatrix}.$$
Solving, we get $x = \begin{pmatrix} \frac{11}{45} \\ -\frac{1}{9} \\ -\frac{14}{45}\end{pmatrix}$.
Now, let's compute $\frac{11}{45}\begin{pmatrix}2 \\ 5 \\4\end{pmatrix} - \frac{1}{9}\begin{pmatrix} 1 \\ 2 \\ -3\end{pmatrix} - \frac{14}{45}\begin{pmatrix} -2\\0\\1\end{pmatrix}$:
$$\color{blue}{\frac{11}{45}\begin{pmatrix}2 \\ 5 \\4\end{pmatrix}}  \color{red}{-\frac{1}{9}\begin{pmatrix} 1 \\ 2 \\ -3\end{pmatrix} - \frac{14}{45}\begin{pmatrix} -2\\0\\1\end{pmatrix}} = \begin{pmatrix} \frac{22}{45} - \frac{5}{45} + \frac{28}{45} \\ \frac{55}{45} - \frac{10}{45} - \frac{0}{45} \\
\frac{44}{45} + \frac{15}{45} - \frac{14}{45} \end{pmatrix} = \begin{pmatrix} \frac{45}{45} \\ \frac{45}{45} \\ \frac{45}{45} \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.$$
Now, the term in $\color{blue}{\textrm{blue}}$ is a vector in $V^\perp$, and the term in $\color{red}{\textrm{red}}$ is a linear combination of vectors in $V$. Since $V$ is a subspace, it is algebraically closed under linear combinations of its elements; hence, the red term is in $V$.
Therefore, let $v$ be the $\color{red}{\textrm{red}}$ term, and let $w$ be the $\color{blue}{\textrm{blue}}$ term, and we are done.
